Question

Sum of the two digits of a number is $9$.

When we interchange the digits the new number is $27$ greater than the earlier number.

Find the number.

Answer 1

Solve for the required number:

Let the number be $xy$ and therefore can be written as $10x+y$

Given that,

• Sum of the two digits of a number is $9$ $\Rightarrow x+y=9\Rightarrow x=9-y$
• On interchanging the digits the new number is $27$ greater than the earlier number.

On interchanging the digits the new number is $yx$.

According to the question,

$\begin{array}{crcll}& & & & \\ \Rightarrow & \left(10y+x\right)-\left(10x+y\right)& =& 27& \\ \Rightarrow & 9y-9x& =& 27& \\ \Rightarrow & y-x& =& \frac{27}{9}& \\ \Rightarrow & y-x& =& 3& \\ \Rightarrow & y-\left(9-y\right)& =& 3& \left[\because x+y=9\right]\\ \Rightarrow & y-9+y& =& 3& \\ \Rightarrow & 2y& =& 9+3& \\ \Rightarrow & y& =& \frac{12}{2}& \\ \Rightarrow & y& =& 6& \\ \therefore & x& =& 9-6& \left[\because x=9-y\right]\\ \Rightarrow & x& =& 3& \end{array}$

$\therefore xy=36$

Hence, the number is $36$